Ring of mass 6 kg and radius 40 cm is revolving at the rate of 300 revolutions per minute. Its moment of inertia will be
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I=MR²=6 × 40/100 × 40/100 kg m² = 0.96kg m²
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Given:
- The mass of the ring = 60 Kg
- The radius of the ring = 40 cm = 0.4 m
- Total revolutions per minute = 300
To Find:
- The moment of Inertia of the ring.
Solution:
The formula to find the moment of inertia is given by,
I = M× → {equation 1}
Where "M" is the mass and "R" is the radius.
On substituting the given values in equation 1 we get,
⇒ I = 6×(0.4×0.4) {multiplying the terms in the bracket}
⇒ I = 6×0.16 {multiplying the terms}
⇒ I = 0.96 Kg
∴ The moment of Inertia of the ring = 0.96 Kg
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